l'analyse de xml retourné par la réponse du service web

Vous pouvez analyser xml pour Dictionary<string, string> avec Linq to Xml:

var xdoc = XDocument.Parse(xml);
var items = xdoc.Descendants("Item")
.ToDictionary(i => (string)i.Attribute("Key"),
i => (string)i.Attribute("Value"));

Utilisation:

string driveline = items["Driveline"];

Ou

foreach(var kvp in items)
//use key value pair

Il est une autre façon que j'ai l'habitude de l'utiliser pour traiter avec xml.

Étape 1: Copie de votre chaîne xml et coller comme un classes.

Lorsque vous le collez, ce code sera généré automatiquement:

        ///<remarks/>
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
[System.Xml.Serialization.XmlRootAttribute(Namespace = "", IsNullable = false)]
public partial class VINquery
{
private VINqueryVIN vINField;
private string versionField;
private string dateField;
///<remarks/>
public VINqueryVIN VIN
{
get
{
return this.vINField;
}
set
{
this.vINField = value;
}
}
///<remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string Version
{
get
{
return this.versionField;
}
set
{
this.versionField = value;
}
}
///<remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string Date
{
get
{
return this.dateField;
}
set
{
this.dateField = value;
}
}
}
///<remarks/>
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public partial class VINqueryVIN
{
private VINqueryVINVehicle vehicleField;
private string numberField;
private string statusField;
///<remarks/>
public VINqueryVINVehicle Vehicle
{
get
{
return this.vehicleField;
}
set
{
this.vehicleField = value;
}
}
///<remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string Number
{
get
{
return this.numberField;
}
set
{
this.numberField = value;
}
}
///<remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string Status
{
get
{
return this.statusField;
}
set
{
this.statusField = value;
}
}
}
///<remarks/>
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public partial class VINqueryVINVehicle
{
private VINqueryVINVehicleItem[] itemField;
private ushort vINquery_Vehicle_IDField;
private ushort model_YearField;
private string makeField;
private string modelField;
private string trim_LevelField;
///<remarks/>
[System.Xml.Serialization.XmlElementAttribute("Item")]
public VINqueryVINVehicleItem[] Item
{
get
{
return this.itemField;
}
set
{
this.itemField = value;
}
}
///<remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public ushort VINquery_Vehicle_ID
{
get
{
return this.vINquery_Vehicle_IDField;
}
set
{
this.vINquery_Vehicle_IDField = value;
}
}
///<remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public ushort Model_Year
{
get
{
return this.model_YearField;
}
set
{
this.model_YearField = value;
}
}
///<remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string Make
{
get
{
return this.makeField;
}
set
{
this.makeField = value;
}
}
///<remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string Model
{
get
{
return this.modelField;
}
set
{
this.modelField = value;
}
}
///<remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string Trim_Level
{
get
{
return this.trim_LevelField;
}
set
{
this.trim_LevelField = value;
}
}
}
///<remarks/>
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public partial class VINqueryVINVehicleItem
{
private string keyField;
private string valueField;
private string unitField;
///<remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string Key
{
get
{
return this.keyField;
}
set
{
this.keyField = value;
}
}
///<remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string Value
{
get
{
return this.valueField;
}
set
{
this.valueField = value;
}
}
///<remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string Unit
{
get
{
return this.unitField;
}
set
{
this.unitField = value;
}
}
}

Étape 2:
Désérialiser votre chaîne xml pour cette classe générée.

public T DeserializeXML<T>(string xmlContent)
{
XmlSerializer serializer = new XmlSerializer(typeof(T));
MemoryStream memStream = new MemoryStream(Encoding.UTF8.GetBytes(xmlContent));
return (T)serializer.Deserialize(memStream);
}
....
string resultXml = resp.GetResponseStream();
VINquery resultObject = DeserializeXML<VINquery>(resultObject); 

Maintenant, vous pouvez facilement manipuler et filtrer vos résultats.

P. S.: Vous devez supprimer les caractères spéciaux (&) dans votre chaîne xml, car elle provoque un problème lorsque vous essayez de coller en tant que classes.